Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 9x}{x + 4} = \dfrac{19x - 24}{x + 4}$
Multiply both sides by $x + 4$ $ \dfrac{x^2 + 9x}{x + 4} (x + 4) = \dfrac{19x - 24}{x + 4} (x + 4)$ $ x^2 + 9x = 19x - 24$ Subtract $19x - 24$ from both sides: $ x^2 + 9x - (19x - 24) = 19x - 24 - (19x - 24)$ $ x^2 + 9x - 19x + 24 = 0$ $ x^2 - 10x + 24 = 0$ Factor the expression: $ (x - 4)(x - 6) = 0$ Therefore $x = 4$ or $x = 6$ The original expression is defined at $x = 4$ and $x = 6$, so there are no extraneous solutions.